I want to know how to prove this:

$$ \sum_{k=1}^\infty \arctan\left( \frac{x}{k^2+a} \right) = \pi \left\lfloor \frac{b}{\sqrt{2}} \right\rfloor + \arctan\left( \frac{x}{b^2} \right) - \arctan\left[ \tanh\left( \frac{\pi}{\sqrt{2}} \frac{x}{b} \right) \cot\left( \frac{\pi}{\sqrt{2}} b \right) \right] $$

for $0 \le a, b = \sqrt{\sqrt{x^2 + a^2} - a}$.

At least in the special case $x=1$,$a=0$

Thank you ..


marked as duplicate by Andrés E. Caicedo, Newb, Michael Albanese, Fixed Point, user66733 Jan 11 '14 at 9:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Let $\;\displaystyle u + iv = \pi\sqrt{a+ix} = \frac{\pi}{\sqrt{2}}\left(\frac{x}{b} + ib\right)$ and follow nearly exactly the same step as in this answer, you get

$$\begin{align} \sum_{k=1}\tan^{-1}\left(\frac{x}{k^2+a}\right) &= \tan^{-1}\left(\frac{\tan v}{\tanh u}\right) - \tan^{-1}\left(\frac{v}{u}\right) + \pi N\\ &= \tan^{-1}\left(\frac{x}{b^2}\right) - \tan^{-1}\left(\tanh\left(\frac{\pi x}{\sqrt{2}{b}}\right)\cot\left(\frac{\pi b}{\sqrt{2}}\right)\right) + \pi N \end{align}$$ for some integer $N$. The only thing left is to figure out what $N$ is...


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